1000 Solved Problems in Modern Physics

1.3 Solutions 39

1.3.2 FourierSeriesandFourierTransforms.................

1.17 f(x)=^1
2

a 0 +

∑∞

n= 1

(

ancos

(nπx

L

)

+bnsin

(nπx

L

))

(1)

an=(1/L)

∫L

−L

f(x) cos

(nπx

L

)

dx (2)

bn=(1/L)

∫L

−L

f(x)sin

(nπx

L

)

dx (3)

Asf(x) is an odd function,an=0 for alln.

bn=(1/L)

∫L

−L

f(x)sin

(nπx

L

)

dx

=(2/L)

∫L

0

xsin

(nπx

L

)

dx

=−

(

2

nπ

)

cosnπ=−

(

2

nπ

)

(−1)n=

(

2

nπ

)

(−1)n+^1

Therefore,

f(x)=(2/π)

∑∞

1

(−1)n+^1

n

sin

(nπx

L

)

=(2/π)[sin

(πx

L

)

−

1

2

sin

(

2 πx

L

)

+

(

1

3

)

sin

(

3 πx

L

)

−···]

Figure 1.7 shows the result for first 3 terms, 6 terms and 9 terms of the

Fourier expansion. As the number of terms increases, a better agreement with

the function is reached. As a general rule if the original function is smoother

compared to, say the saw-tooth function the convergence of the Fourier series

is much rapid and only a few terms are required. On the other hand, a highly

discontinuous function can be approximated with reasonable accuracy only

with large number of terms.

Fig. 1.7Fourier expansion of the saw-tooth wave